Solve (x-y)(x+y)-(3x-y)(3x+y)

$$( x - y ) ( x + y ) - ( 3 x - y ) ( 3 x + y )$$

Evaluate

$$-8x^{2}$$

Solution Steps

Consider \left(x-y\right)\left(x+y\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$x^{2}-y^{2}-\left(3x-y\right)\left(3x+y\right)$$

Consider \left(3x-y\right)\left(3x+y\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$x^{2}-y^{2}-\left(\left(3x\right)^{2}-y^{2}\right)$$

Expand \left(3x\right)^{2}.

$$x^{2}-y^{2}-\left(3^{2}x^{2}-y^{2}\right)$$

Calculate 3 to the power of 2 and get 9.

$$x^{2}-y^{2}-\left(9x^{2}-y^{2}\right)$$

To find the opposite of 9x^{2}-y^{2}, find the opposite of each term.

$$x^{2}-y^{2}-9x^{2}-\left(-y^{2}\right)$$

The opposite of -y^{2} is y^{2}.

$$x^{2}-y^{2}-9x^{2}+y^{2}$$

Combine x^{2} and -9x^{2} to get -8x^{2}.

$$-8x^{2}-y^{2}+y^{2}$$

Combine -y^{2} and y^{2} to get 0.

$$-8x^{2}$$

Expand

$$-8x^{2}$$

Solution Steps

Consider \left(x-y\right)\left(x+y\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$x^{2}-y^{2}-\left(3x-y\right)\left(3x+y\right)$$

Consider \left(3x-y\right)\left(3x+y\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$x^{2}-y^{2}-\left(\left(3x\right)^{2}-y^{2}\right)$$

Expand \left(3x\right)^{2}.

$$x^{2}-y^{2}-\left(3^{2}x^{2}-y^{2}\right)$$

Calculate 3 to the power of 2 and get 9.

$$x^{2}-y^{2}-\left(9x^{2}-y^{2}\right)$$

To find the opposite of 9x^{2}-y^{2}, find the opposite of each term.

$$x^{2}-y^{2}-9x^{2}-\left(-y^{2}\right)$$

The opposite of -y^{2} is y^{2}.

$$x^{2}-y^{2}-9x^{2}+y^{2}$$

Combine x^{2} and -9x^{2} to get -8x^{2}.

$$-8x^{2}-y^{2}+y^{2}$$

Combine -y^{2} and y^{2} to get 0.

$$-8x^{2}$$